If we knew Taylor series of $f(x)$ is it possible to get Taylor series of $f^{-1}(x)$? If we knew Taylor series of $f(x)$ is it possible to get Taylor series of $f^{-1}(x)$?
Suppose,

$$f(x)=a_0+a_1x+a_2x^2+\cdots$$

Then

$$f^{-1}(x)=?$$

If possible,how do we get it?
Easiest would be to expand at $x=a_0$
 A: What we want is all possible derivatives of $f^{-1}$ at $x = a_0$. We can get those from using that $f(f^{-1}(x)) = x$ and using the chain rule. First the constant term is clearly $0$, since $f^{-1}(a_0) = 0$. Differentiating once, we get
$$
[f(f^{-1}(x))]' = [x]'\\
{f^{-1}}'(x)f'(f^{-1}(x)) = 1\\
{f^{-1}}'(a_0)f'(f^{-1}(a_0)) = 1\\
{f^{-1}}'(a_0) = \frac1{f'(0)} = \frac1{a_1}
$$
For the second term, we differentiate once more:
$$
[{f^{-1}}'(x)f'(f^{-1}(x))]' = [1]'\\
{f^{-1}}''(x)f'(f^{-1}(x)) + ({f^{-1}}'(x))^2f''(f^{-1}(x)) = 0\\
{f^{-1}}''(a_0)f'(f^{-1}(a_0)) + ({f^{-1}}'(a_0))^2f''(f^{-1}(a_0)) = 0\\
{f^{-1}}''(a_0) = -\frac{\frac{1}{a_1^2}\cdot 2a_2}{a_1} = -\frac{2a_2}{a_1^3}
$$
and we can keep going, at least in theory, getting
$$
f^{-1}(x) = \frac1{a_1}(x-a_0) - \frac{a_2}{a_1^3}(x-a_0)^2 + \cdots
$$
So, assuming the Taylor series for the inverse exists (which so far amounts to $a_1\neq 0$, and since ${f^{-1}}^{(n)}(x)$ in the $n$'th differentiation always appears only together with $f'(f^{-1}(x))$ as a factor, $a_1 = 0$ seems to me like the only thing that would ever go wrong), there is at least one way to find all the coefficients. I'm sure there are more efficient ways, or that someone else has found a general form for these terms (or at the very least carried this calculation further than the second degree).
A: Let $f(x)=x^3$, then the Taylor series of $f$ around $x_=0$ is
$f(x)=0+0 \cdot x+0 \cdot x^2+1 \cdot x^3+0 \cdot x^4 +....$.
But $f^{-1}$ is not differentiable at $x_0=0$, hence the Taylor series of $f^{-1}$ does not exist !
A: Let be Taylor expansion of $f$ near $a\in R$, that is
$$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+R_{n+1}(x)$$
Next, let find $f^{-1}(x)$, to do it recll multinomial theorem for case $n=-1$,
$$f^{-1}(x)=\left(f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+R_{n+1}(x)\right)^{-1}$$
How to use multinomial theorem in negative exponents is shonw in this discussion
Wikipaedia article is here
